Is ${305650}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {305650}= &&{3}\cdot100000+ \\&&{0}\cdot10000+ \\&&{5}\cdot1000+ \\&&{6}\cdot100+ \\&&{5}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {305650}= &&{3}(99999+1)+ \\&&{0}(9999+1)+ \\&&{5}(999+1)+ \\&&{6}(99+1)+ \\&&{5}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {305650}= &&\gray{3\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {3}+{0}+{5}+{6}+{5}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${305650}$ is divisible by $3$ if ${ 3}+{0}+{5}+{6}+{5}+{0}$ is divisible by $3$ Add the digits of ${305650}$ $ {3}+{0}+{5}+{6}+{5}+{0} = {19} $ If ${19}$ is divisible by $3$ , then ${305650}$ must also be divisible by $3$ ${19}$ is not divisible by $3$, therefore ${305650}$ must not be divisible by $3$.